Is this proof for a sequence convergence problem correct?

[jgens]

Homework Statement

\lim_{n\to\infty}a_n=l \rightarrow \lim_{n\to\infty}\frac{a_1+\dots+a_n}{n}=l

Homework Equations

N/A

The Attempt at a Solution

Could someone verify that this proof works? I would really appreciate it.

Proof: Since the sequence \{a_n\} converges to l, for any given \varepsilon>0 it's possible to find a number N>0 such that if n>N, then |a_n-l|<\varepsilon/2. Now, because there are only finitely many numbers |a_1-l|,\dots,|a_N-l|, we can choose the greatest such number. Denote this number by M.

Suppose that n>\max{(N,\frac{2MN}{\varepsilon})}, in which case, it clearly follows that \frac{\varepsilon}{2}>\frac{MN}{n}. Therefore,

\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|\leq\frac{|a_1-l|}{n}+\dots+\frac{|a_N-l|}{n}\leq\frac{MN}{n}<\frac{\varepsilon}{2}

Moreover, since n>N, we also have that

\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|\leq\frac{|a_{N+1}-l|}{n}+\dots+\frac{|a_n-l|}{n}<\frac{(n-N)\varepsilon}{2n}<\frac{\varepsilon}{2}

Combining these two results, we see that

\left| \frac{a_1+\dots+a_n}{n}-l\right|\leq\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|+\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon

Completing the proof.

 

[jgens]

Bump.

 

[snipez90]

Looks good to me.

 

[jgens]

Thanks!